The cases are (2), (3), (4), and (5) women. The total is \(\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752\).
Step 2
Why this answer is correct
The correct answer is D. (3752). The cases are (2), (3), (4), and (5) women. The total is \(\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752\).
Step 3
Exam Tip
मामले (2), (3), (4) और (5) महिलाओं के हैं। कुल \(\binom{8}{2}\binom{8}{3}+\binom{8}{3}\binom{8}{2}+\binom{8}{4}\binom{8}{1}+\binom{8}{5}=3752\) है।
The cases are choosing (2), (3), or (4) from the first (4). The total is \(\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371\).
Step 2
Why this answer is correct
The correct answer is C. (371). The cases are choosing (2), (3), or (4) from the first (4). The total is \(\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371\).
Step 3
Exam Tip
मामले पहले (4) में से (2), (3) या (4) चुनने के हैं। कुल \(\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2}=371\) है।
Total ways are \(\binom{13}{5}=1287\) and ways with both special books are \(\binom{11}{3}=165\). Hence (1287-165=1122).
Step 2
Why this answer is correct
The correct answer is B. (1188). Total ways are \(\binom{13}{5}=1287\) and ways with both special books are \(\binom{11}{3}=165\). Hence (1287-165=1122).
Step 3
Exam Tip
कुल \(\binom{13}{5}=1287\) हैं और दोनों विशेष साथ हों तो \(\binom{11}{3}=165\) हैं। इसलिए (1287-165=1122) तरीके हैं।
Total pairs are \(\binom{18}{2}=153\) and (7) collinear points give (1) line instead of \(\binom{7}{2}\). Hence (153-21+1=133).
Step 2
Why this answer is correct
The correct answer is B. (133). Total pairs are \(\binom{18}{2}=153\) and (7) collinear points give (1) line instead of \(\binom{7}{2}\). Hence (153-21+1=133).
Step 3
Exam Tip
कुल \(\binom{18}{2}=153\) जोड़ियां हैं और (7) समरेखीय बिंदु \(\binom{7}{2}\) के स्थान पर (1) रेखा देते हैं। इसलिए (153-21+1=133) है।
Total triples are \(\binom{14}{3}=364\) and \(\binom{6}{3}=20\) collinear triples do not form triangles. Hence (364-20=344).
Step 2
Why this answer is correct
The correct answer is A. (344). Total triples are \(\binom{14}{3}=364\) and \(\binom{6}{3}=20\) collinear triples do not form triangles. Hence (364-20=344).
Step 3
Exam Tip
कुल \(\binom{14}{3}=364\) त्रिक हैं और \(\binom{6}{3}=20\) समरेखीय त्रिक त्रिभुज नहीं बनाते। इसलिए (364-20=344) है।
The cases are (3), (4), and (5) chemistry books. The total is \(\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932\).
Step 2
Why this answer is correct
The correct answer is C. (1932). The cases are (3), (4), and (5) chemistry books. The total is \(\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932\).
Step 3
Exam Tip
मामले (3), (4) और (5) रसायन पुस्तकों के हैं। कुल \(\binom{9}{3}\binom{6}{2}+\binom{9}{4}\binom{6}{1}+\binom{9}{5}=1932\) है।
Total ways are \(\binom{16}{5}=4368\). Removing only English \(\binom{7}{5}=21\) and only Hindi \(\binom{9}{5}=126\) gives (4221) ways.
Step 2
Why this answer is correct
The correct answer is B. (4326). Total ways are \(\binom{16}{5}=4368\). Removing only English \(\binom{7}{5}=21\) and only Hindi \(\binom{9}{5}=126\) gives (4221) ways.
Step 3
Exam Tip
कुल \(\binom{16}{5}=4368\) हैं। केवल अंग्रेजी \(\binom{7}{5}=21\) और केवल हिंदी \(\binom{9}{5}=126\) हटाने पर (4221) तरीके मिलते हैं।
The cases are (3), (4), and (5) doctors. The total is \(\binom{8}{3}\binom{6}{2}+\binom{8}{4}\binom{6}{1}+\binom{8}{5}=1176\).
Step 2
Why this answer is correct
The correct answer is A. (1176). The cases are (3), (4), and (5) doctors. The total is \(\binom{8}{3}\binom{6}{2}+\binom{8}{4}\binom{6}{1}+\binom{8}{5}=1176\).
Step 3
Exam Tip
मामले (3), (4) और (5) डॉक्टरों के हैं। कुल \(\binom{8}{3}\binom{6}{2}+\binom{8}{4}\binom{6}{1}+\binom{8}{5}=1176\) है।
Total ways are \(\binom{14}{6}=3003\) and ways with no special student are \(\binom{10}{6}=210\). Hence (3003-210=2793) ways.
Step 2
Why this answer is correct
The correct answer is A. (2877). Total ways are \(\binom{14}{6}=3003\) and ways with no special student are \(\binom{10}{6}=210\). Hence (3003-210=2793) ways.
Step 3
Exam Tip
कुल \(\binom{14}{6}=3003\) हैं और कोई विशेष न हो तो \(\binom{10}{6}=210\) हैं। इसलिए (3003-210=2793) तरीके हैं।
Total ways are \(\binom{15}{5}=3003\). Removing only pens \(\binom{8}{5}=56\) and only pencils \(\binom{7}{5}=21\) gives (2926).
Step 2
Why this answer is correct
The correct answer is C. (2982). Total ways are \(\binom{15}{5}=3003\). Removing only pens \(\binom{8}{5}=56\) and only pencils \(\binom{7}{5}=21\) gives (2926).
Step 3
Exam Tip
कुल \(\binom{15}{5}=3003\) हैं। केवल पेन \(\binom{8}{5}=56\) और केवल पेंसिल \(\binom{7}{5}=21\) हटाने पर (2926) है।
By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{11}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{11}{5}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{11}{5}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{11}{5}\) है।
The element (1) is already chosen so the remaining (4) elements are chosen from (9). The number of ways is \(\binom{9}{4}=126\).
Step 2
Why this answer is correct
The correct answer is B. (126). The element (1) is already chosen so the remaining (4) elements are chosen from (9). The number of ways is \(\binom{9}{4}=126\).
Step 3
Exam Tip
(1) पहले से चुना है इसलिए बाकी (4) तत्व (9) में से चुने जाएंगे। तरीकों की संख्या \(\binom{9}{4}=126\) है।
Total subsets are \(\binom{10}{5}=252\) and those containing both (2), (3) are \(\binom{8}{3}=56\). Hence (252-56=196).
Step 2
Why this answer is correct
The correct answer is C. (196). Total subsets are \(\binom{10}{5}=252\) and those containing both (2), (3) are \(\binom{8}{3}=56\). Hence (252-56=196).
Step 3
Exam Tip
कुल \(\binom{10}{5}=252\) हैं और (2), (3) दोनों हों तो \(\binom{8}{3}=56\) हैं। इसलिए (252-56=196) है।
Total ways are \(\binom{18}{6}=18564\). Removing the cases of (0) teacher and (1) teacher gives \(18564-\binom{10}{6}-\binom{8}{1}\binom{10}{5}=16338\).
Step 2
Why this answer is correct
The correct answer is B. (17472). Total ways are \(\binom{18}{6}=18564\). Removing the cases of (0) teacher and (1) teacher gives \(18564-\binom{10}{6}-\binom{8}{1}\binom{10}{5}=16338\).
Step 3
Exam Tip
कुल \(\binom{18}{6}=18564\) हैं। (0) शिक्षक और (1) शिक्षक के मामले हटाने पर \(18564-\binom{10}{6}-\binom{8}{1}\binom{10}{5}=16338\) है।
Total ways are \(\binom{15}{5}=3003\). Removing all-white \(\binom{7}{5}=21\) and all-black \(\binom{8}{5}=56\) gives (2926).
Step 2
Why this answer is correct
The correct answer is A. (2926). Total ways are \(\binom{15}{5}=3003\). Removing all-white \(\binom{7}{5}=21\) and all-black \(\binom{8}{5}=56\) gives (2926).
Step 3
Exam Tip
कुल \(\binom{15}{5}=3003\) हैं। सभी सफेद \(\binom{7}{5}=21\) और सभी काली \(\binom{8}{5}=56\) हटाने पर (2926) मिलते हैं।
Total ways are \(\binom{10}{5}=252\) and if both special friends do not go then \(\binom{8}{5}=56\). Hence (252-56=196).
Step 2
Why this answer is correct
The correct answer is B. (182). Total ways are \(\binom{10}{5}=252\) and if both special friends do not go then \(\binom{8}{5}=56\). Hence (252-56=196).
Step 3
Exam Tip
कुल \(\binom{10}{5}=252\) हैं और दोनों विशेष न जाएं तो \(\binom{8}{5}=56\) हैं। इसलिए (252-56=196) है।
The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (7). The ways are \(\binom{7}{3}=35\).
Step 2
Why this answer is correct
The correct answer is C. (35). The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (7). The ways are \(\binom{7}{3}=35\).
Step 3
Exam Tip
(a) तय है और (b) हट गया है इसलिए बाकी (3) अक्षर (7) में से चुनेंगे। तरीके \(\binom{7}{3}=35\) हैं।
Total ways are \(\binom{8}{4}=70\) and if both special subjects are included there are \(\binom{6}{2}=15\) ways. Hence (70-15=55).
Step 2
Why this answer is correct
The correct answer is C. (55). Total ways are \(\binom{8}{4}=70\) and if both special subjects are included there are \(\binom{6}{2}=15\) ways. Hence (70-15=55).
Step 3
Exam Tip
कुल \(\binom{8}{4}=70\) हैं और दोनों विशेष विषय साथ हों तो \(\binom{6}{2}=15\) हैं। इसलिए (70-15=55) है।
The (2) numbers are fixed and (2) are excluded. The remaining (3) numbers are chosen from (9) in \(\binom{9}{3}=84\) ways.
Step 2
Why this answer is correct
The correct answer is A. (84). The (2) numbers are fixed and (2) are excluded. The remaining (3) numbers are chosen from (9) in \(\binom{9}{3}=84\) ways.
Step 3
Exam Tip
(2) संख्याएं तय हैं और (2) हट गई हैं। बाकी (3) संख्याएं (9) में से \(\binom{9}{3}=84\) तरीकों से चुनी जाएंगी।
For a failed triangle (3) points are chosen from the (7) collinear points. So the number is \(\binom{7}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\binom{7}{3}\). For a failed triangle (3) points are chosen from the (7) collinear points. So the number is \(\binom{7}{3}\).
Step 3
Exam Tip
असफल त्रिभुज के लिए (7) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{7}{3}\) है।
If both are included there are \(\binom{9}{3}=84\) ways and if both are excluded there are \(\binom{9}{5}=126\) ways. The total is (210).
Step 2
Why this answer is correct
The correct answer is C. (280). If both are included there are \(\binom{9}{3}=84\) ways and if both are excluded there are \(\binom{9}{5}=126\) ways. The total is (210).
Step 3
Exam Tip
दोनों शामिल हों तो \(\binom{9}{3}=84\) और दोनों बाहर हों तो \(\binom{9}{5}=126\) तरीके हैं। कुल (84+126=210) है।
The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185\).
Step 2
Why this answer is correct
The correct answer is C. (185). The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185\).
Step 3
Exam Tip
मामले अंतिम (4) में से (2), (3) या (4) चुनने के हैं। कुल \(\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185\) है।
The number of seniors can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772\).
Step 2
Why this answer is correct
The correct answer is B. (2016). The number of seniors can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0), (1) या (2) हो सकती है। कुल \(\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772\) है।
Choose (1) of the two fixed cards and (4) cards from the remaining (10). The ways are \(\binom{2}{1}\binom{10}{4}=420\).
Step 2
Why this answer is correct
The correct answer is B. (420). Choose (1) of the two fixed cards and (4) cards from the remaining (10). The ways are \(\binom{2}{1}\binom{10}{4}=420\).
Step 3
Exam Tip
दो निश्चित कार्डों में से (1) चुनें और बाकी (4) कार्ड (10) में से चुनें। तरीके \(\binom{2}{1}\binom{10}{4}=420\) हैं।
Either (0) or (1) special object can be chosen. The total is \(\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756\).
Step 2
Why this answer is correct
The correct answer is C. (1512). Either (0) or (1) special object can be chosen. The total is \(\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756\).
Step 3
Exam Tip
विशेष वस्तुओं में से (0) या (1) चुनी जा सकती है। कुल \(\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756\) है।
